Given the metric g_{\mu\nu}(X), show that the Euler-Lagrange equation (16) (we drop the “s”), to minimize the action along a trajectory in space-time,
\frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot X^m} = \frac{\partial \mathcal L}{\partial X^m}
where the Lagrangian \mathcal L is
\mathcal L = -m \sqrt{- g_{\mu\nu}(X) \frac{dX^\mu}{dt} \frac{dX^\nu}{dt}}
is equivalent to the definition of a geodesic given by equation (6), which says that the tangent vector to the trajectory in space-time stays constant:
\frac{d^2 X^\mu}{d\tau^2} = - {\Gamma^\mu}_{\sigma\rho}\frac{\mathrm dX^\sigma}{\mathrm d\tau}\frac{dX^\rho}{d\tau}
The Lagrangian is given (we simplify writing g_{\mu\nu}(X) as g_{\mu\nu}):
\mathcal L = -m \sqrt{- g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm dt}} \equiv -m \sqrt{F}
The partial derivative with respect to velocity \dot{X}^m is:
\begin{aligned} \frac{\partial \mathcal L}{\partial \dot X^m} &=\frac{\partial}{\partial \dot X^m}\left(-m\sqrt F \right) = \left(\frac{-m}{2 \sqrt F}\right)\frac{\partial F}{\partial \dot X^m} \\ & = -\frac{m}{2 \sqrt F}\frac{\partial }{\partial \dot X^m} \left(- g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm dt}\right) \\ & = -\frac{m}{2 \sqrt F}\frac{\partial }{\partial \dot X^m} \left(- g_{\mu\nu} \dot X^\mu \dot X^\nu\right) \\ & = -\frac{m}{2 \sqrt F} (- g_{\mu\nu}) \left(\frac{\partial \dot X^\mu}{\partial \dot X^m} \dot X^\nu + \frac{\partial \dot X^\nu}{\partial \dot X^m} \dot X^\mu \right) \\ & = \frac{m}{2 \sqrt F}g_{\mu\nu} \left({\delta^\mu}_m \dot X^\nu + {\delta^\nu}_m \dot X^\mu \right) \\ & = \frac{m}{2 \sqrt F}g_{\mu\nu} \left({\delta^\mu}_m \dot X^\nu + {\delta^\nu}_m \dot X^\mu \right) \\ & = \frac{m}{2 \sqrt F}\left(g_{m\nu} \dot X^\nu + g_{\mu m} \dot X^\mu \right) \\ & = \frac{m}{2 \sqrt F}\left(2 g_{m\nu} \dot X^\nu \right) \\ & = \frac{m g_{m\nu}}{\sqrt F} \frac{\mathrm dX^\nu}{\mathrm dt} \end{aligned}
From:
\mathrm d\tau^2 = - g_{\mu\nu} \mathrm dX^\mu \mathrm dX^\nu
We have:
\frac{\mathrm d\tau}{\mathrm dt} = \sqrt {\frac{\mathrm d\tau^2}{\mathrm dt^2}} = \sqrt{- g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm dt}} = \sqrt F
Substituting:
\frac{\partial \mathcal L}{\partial \dot X^m} = \frac{m g_{m\nu}}{\sqrt F} \frac{\mathrm dX^\nu}{\mathrm dt} = m g_{m\nu}\frac{\frac{\mathrm dX^\nu}{\mathrm dt}}{\frac{\mathrm d\tau}{\mathrm dt}} = m g_{m\nu}\frac{\mathrm dX^\nu}{\mathrm d\tau}
Now we can compute the full left-hand side of the Euler-Lagrange equation by taking the total derivative with respect to t:
\begin{aligned} \frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal L}{\partial \dot X^m} \right) &= m \frac{\mathrm d}{\mathrm dt} \left( g_{m\nu} \frac{\mathrm dX^\nu}{\mathrm d\tau} \right) \\ &= m \left[ \left( \frac{\mathrm d g_{m\nu}}{\mathrm dt} \right) \frac{\mathrm dX^\nu}{\mathrm d\tau} + g_{m\nu} \frac{\mathrm d}{\mathrm dt} \left( \frac{\mathrm dX^\nu}{\mathrm d\tau} \right) \right] \end{aligned}
We can expand these total derivatives using the chain rule, noting that g_{m\nu} depends on position X^\sigma, which depends on t:
\frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal L}{\partial \dot X^m} \right) = m \left(\partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm d\tau} + g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} \frac{\mathrm d\tau}{\mathrm dt} \right)
Next, we compute the right-hand side, \frac{\partial \mathcal L}{\partial X^m}. The dependence on X^m is entirely within the metric tensor:
\begin{aligned} \frac{\partial \mathcal L}{\partial X^m} &= \frac{\partial}{\partial X^m}\left(-m\sqrt F \right) = \left(\frac{-m}{2 F}\right)\frac{\partial \sqrt F}{\partial X^m}\\ & = \left(\frac{m}{2 \sqrt F}\right)\partial_m g_{\mu\nu} \dot{X}^\mu \dot{X}^\nu \end{aligned}
Again, we substitute \sqrt F = \frac{\mathrm d\tau}{\mathrm dt} for the denominator and also re-express the velocities \dot{X}^\mu in terms of proper time (\dot{X}^\mu = \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt}):
\begin{aligned} \frac{\partial \mathcal L}{\partial X^m} &= \frac{m}{2 \frac{\mathrm d\tau}{\mathrm dt}} \partial_m g_{\mu\nu} \left( \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt} \right) \left( \frac{\mathrm dX^\nu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt} \right) \\ &= \frac{m}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt} \end{aligned}
Now we equate the left and right sides of the Euler-Lagrange equation and cancel the common factor of m:
\partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm d\tau} + g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} \frac{\mathrm d\tau}{\mathrm dt} = \frac{1}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt}
To simplify, we multiply the entire equation by \frac{\mathrm dt}{\mathrm d\tau} and use \frac{\mathrm dX^\sigma}{\mathrm dt} = \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt}.
\partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} + g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = \frac{1}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau}
Isolating the second derivative term gives:
g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = \frac{1}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} - \partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau}
We can manipulate the indices:
\begin{aligned} g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} & = \frac{1}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} - \partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau}\\ & = \frac{1}{2} \partial_m g_{\sigma\nu} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} - \partial_\sigma g_{m\rho} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau}\\ & = \frac{1}{2} \partial_m g_{\sigma\rho} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} - \partial_\sigma g_{m\rho} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau}\\ & = \left( \frac{1}{2} \partial_m g_{\sigma\rho} - \partial_\sigma g_{m\rho} \right) \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} \\ &= \frac{1}{2} \left( \partial_m g_{\sigma\rho} - \partial_\sigma g_{m\rho} - \partial_\sigma g_{m\rho} \right) \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} \end{aligned}
We have -\partial_\sigma g_{m\rho} = \frac{1}{2}\left(- \partial_\sigma g_{m\rho} - \partial_\sigma g_{m\rho} \right) that can be written as (just changing the indices):
\partial_\sigma g_{m\rho} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} = \partial_a g_{m b} \frac{\mathrm dX^a}{\mathrm d\tau} \frac{\mathrm dX^b}{\mathrm d\tau} = \partial_\rho g_{m \sigma} \frac{\mathrm dX^\rho}{\mathrm d\tau} \frac{\mathrm dX^\sigma}{\mathrm d\tau} = \partial_\rho g_{m \sigma} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau}
So:
g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = \frac{1}{2} \left(\partial_m g_{\sigma\rho} - \partial_\sigma g_{m\rho} -\partial_\rho g_{m\sigma} \right) \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau}
Finally, we multiply by the inverse metric g^{\mu m}:
\begin{aligned} & g^{\mu m} g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = {\delta^\mu}_\nu\frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = \frac{\mathrm d^2 X^\mu}{\mathrm d\tau^2} \\ & = \frac{1}{2} g^{\mu m} \left(\partial_m g_{\sigma\rho} - \partial_\sigma g_{m\rho} -\partial_\rho g_{m\sigma} \right) \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} \end{aligned}
The term on the right involving the metric derivatives is precisely the definition of the Christoffel symbol {\Gamma^\mu}_{\sigma\rho}, with a negative sign:
{\Gamma^\mu}_{\sigma\rho} = \frac{1}{2} g^{\mu m} \left(\partial_\sigma g_{m\rho} + \partial_\rho g_{m\sigma} - \partial_m g_{\sigma\rho} \right)
This leaves us with the final expression:
\frac{\mathrm d^2 X^\mu}{\mathrm d\tau^2} = - {\Gamma^\mu}_{\sigma\rho}\frac{\mathrm dX^\sigma}{\mathrm d\tau}\frac{\mathrm dX^\rho}{\mathrm d\tau}
This result confirms that the path of least action for a free particle in a curved spacetime is the geodesic.
Finally, when we use this Lagrangian, the Euler-Lagrange equations will of course simply produce Newton’s equation for a particle in a gravitational field U(X), just as it did when we carried out exactly the same calculations in volume 1.
We arrive at
m \ddot X = -m \frac{\partial U}{\partial X}
The approximated action is:
A = \int \left(mc^2 - mU(X) + \frac{m}{2}\dot X^2\right) \mathrm dt
We can recognize the Lagrangian in the form kinetic energy minus potential energy (with a constant term mc^2 which is constant and therefore is not entering in the equation):
\mathcal L = T - U
We can apply the Euler-Lagrange equation:
\frac{\mathrm d}{\mathrm dt} \left(\frac{\partial \mathcal L}{\partial \dot X^m}\right) = \frac{\partial \mathcal L}{\partial X^m}
For the left hand side, there is only one term that depends from \dot X:
\frac{\mathrm d}{\mathrm dt} \left(\frac{\partial \mathcal L}{\partial \dot X^m}\right) = \frac{\mathrm d}{\mathrm dt} \left(m \dot X\right) = m \ddot X
For the right hand side, there is only one term that depends from X:
\frac{\partial \mathcal L}{\partial X^m} = -m \frac{\partial U(X)}{\partial X}
Putting all together we derive the equation of motions:
m \ddot X = -m \frac{\partial U(X)}{\partial X}
This is the standard Newton equation for a particle in a gravitational field U(X).
Since there is a mass on both side, it cancels out:
\ddot X = -\frac{\partial U(X)}{\partial X}
so the motion does not depends from the mass of the object.